Factoring (four polynomials)

Text

Factor: (a) $162c^4a^6-128c^4b^4$; (b) $ab+3a-2b-6$; (c) $d^2-5d-14$; (d) $32a^2c-128b^2c^3$.

Solution

(a) Factor out the common term $2c^4$: $162c^4a^6-128c^4b^4=2c^4(81a^6-64b^4).$ Now $81a^6-64b^4=(9a^3)^2-(8b^2)^2$ is a difference of squares: $2c^4(81a^6-64b^4)=2c^4(9a^3-8b^2)(9a^3+8b^2).$

(b) Grouping four terms in pairs: $ab+3a-2b-6=a(b+3)-2(b+3).$ Now $(b+3)$ is a common factor: $a(b+3)-2(b+3)=(b+3)(a-2).$

(c) Characteristic trinomial: two numbers with sum $-5$ and product $-14$, namely $-7$ and $2$. $d^2-5d-14=(d-7)(d+2).$

(d) Factor out $32c$: $32a^2c-128b^2c^3=32c(a^2-4b^2c^2).$ Now $a^2-4b^2c^2=a^2-(2bc)^2$ is a difference of squares: $32c(a^2-4b^2c^2)=32c(a-2bc)(a+2bc).$

$\boxed{2c^4(9a^3-8b^2)(9a^3+8b^2)\ ;\ \ (b+3)(a-2)\ ;\ \ (d-7)(d+2)\ ;\ \ 32c(a-2bc)(a+2bc)}$